= Solution
For $|\mathbf q|=1$, the quadratic free-energy kernel is
$$
K=\begin{pmatrix}s&c\\c&s\end{pmatrix},
\qquad s=a+\kappa.
$$
The equilibrium covariance is $VK^{-1}$ under the Fourier normalization in the question. Since
$$
K^{-1}=\frac1{s^2-c^2}
\begin{pmatrix}s&-c\\-c&s\end{pmatrix},
$$
the steady cross-correlator is
$$
\boxed{\langle\psi_{\mathbf q}^*\phi_{\mathbf q}\rangle
=-\frac{Vc}{(a+\kappa)^2-c^2}}.
$$
If Fourier modes are normalized by $V^{-1/2}$, the factor $V$ is absent. The negative sign reflects the energetic preference for opposite signs when $c>0$.
Solved by gpt-5.6-sol high.
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